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It's given that $a \in \mathbb{R}$ with $a > 0$ and $0<x_1<\frac{1}{a}$. The sequence $(x_n)_{n\in\mathbb{N}}$ is defined by $x_{n+1} = 2x_n-ax^2_n.$ Now, I first proved by induction that $x_n < \frac{1}{n}$ for all $n \in \mathbb{N}$. Then I was wondering, if it is enough to just show that $x_{n+1} = 2x_n-ax^2_n > 2x_n - \frac{1}{x_n}x^2_n = 2x_n-x_n = x_n$ to fully prove that $ x_{n+1} > x_n$ for all $n \in \mathbb{N}$ and therefore $(x_n)_{n\in\mathbb{N}}$ is monotonically increasing? Thanks for any corrections or tips, I'm just not sure if I might be missing something.

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    $\begingroup$ Yes, it would be enough, but how do you prove it? $\endgroup$ – Bernard Jul 10 '19 at 22:58
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    $\begingroup$ How do you know that $\frac1{x_n}\gt a$? $\endgroup$ – Chris Custer Jul 10 '19 at 22:59
  • $\begingroup$ 1.: $x_1 < \frac{1}{a}$ by Definition. 2.: Let's suppose that $x_n < \frac{1}{a}$ is true. Then $x_{n+1} < \frac{1}{a}$ should follow. 3.: $x_{n+1} = 2x_n-ax^2_n = x_n(2-ax_n) < \frac{1}{a}(2-\frac{1}{a}a) = \frac{1}{a}.$ $\endgroup$ – psyph Jul 10 '19 at 23:05
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    $\begingroup$ I don't think your induction step works out, because the second $x_n$ is replaced by the bigger $\frac1a$, and then subtracted. $\endgroup$ – Chris Custer Jul 10 '19 at 23:13
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    $\begingroup$ Your proof is OK but you have to justify the inequality $x_n(2-ax_n) <\frac 1 a (2-\frac 1 a a)$. $\endgroup$ – Kavi Rama Murthy Jul 10 '19 at 23:14
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You have to justify the inequality $x_n(2-ax_n) <\frac 1 a (2-\frac 1 a a)$. To prove this consider the function $f(t)=t(2-at)$ defined on $(0,\frac 1 a)$. Since $f'(t)=2-2at>0$ for all $t$ in the domain of $f$ this function is increasing. Since $t(2-at)=\frac 1 a$ when $t=\frac 1 a$ it follows (by looking at the extension of $f$ to $(0,\frac 1 a]$) that $f(t) <\frac 1 a$ for all $t$. So if we assume that $x_n <\frac 1 a$ we can take $t=x_n$ to see that $x_{n+1}<\frac 1 a$.

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  • $\begingroup$ Thanks, that made it so clear. Now can I also ask you, if I want to show that my sequence converges to $\frac{1}{a}$, can I make my proof by showing that $lim_{n\to\infty} (x_{n+1}-\frac{1}{a})=0$? $\endgroup$ – psyph Jul 10 '19 at 23:41
  • $\begingroup$ The sequence is increasing and bounded above by $\frac 1 a$. This implies that it is convergent. If you call the limit $l$ then the given equation gives $l=2l-al^{2}$ or $l=al^{2}$. This means $l=0$ or $l =\frac 1 a$. Since $x_n >0$ (proved again by induction) for all $n$ and $x_n$ is increasing its limit cannot be $0$. Hence $l =\frac 1 a$. $\endgroup$ – Kavi Rama Murthy Jul 10 '19 at 23:49
  • $\begingroup$ Thanks, but which equations gives $l=2l-al^2$ or $al^2$? And what about the proof that I proposed? Is it sufficient? Because it works out since $lim_{n\to\infty}(x_{n+1}-\frac{1}{a})$ is equal to zero. $\endgroup$ – psyph Jul 10 '19 at 23:54
  • $\begingroup$ How do you show that $x_{n+1}-\frac 1 a \to 0$? $\endgroup$ – Kavi Rama Murthy Jul 10 '19 at 23:58
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    $\begingroup$ I am just taking limit as $ n \to \infty$ in the equation $x_{n+1}=2x_n-ax_n^{2}$. This gives $l=2l-al^{2}$. We can re-write this equation as $al^{2}=l$. This gives $l(al-1)=0$. So either $l=0$ or $al=1$. Once you rule out $l=0$ we are left with $al=1$ which gives $l=\frac 1 a$. $\endgroup$ – Kavi Rama Murthy Jul 11 '19 at 0:09

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